package hDOJ;

/*

There are many lamps in a line. All of them are off at first. A series of operations 
are carried out on these lamps. On the i-th operation, the lamps whose numbers are the 
multiple of i change the condition ( on to off and off to on )
翻译：有很多灯放在一条线上。所有灯一开始都是关闭着的。一连串的操作是执行在这些灯上的。在第i次操作，这些编号为i的
倍数的灯改变状态（开到关或关到开）
Each test case contains only a number n ( 0< n<= 10^5) in a line.
翻译：每一个案例的输入占一行，仅包含一个数字n
Output the condition of the n-th lamp after infinity operations ( 0 - off, 1 - on ).
在无数次的操作之后输出第n个灯的状态（0-关，1-开）
Input
1
5


Output
1
0

Consider the second test case:

The initial condition      : 0 0 0 0 0 …
After the first operation  : 1 1 1 1 1 …
After the second operation : 1 0 1 0 1 …
After the third operation  : 1 0 0 0 1 …
After the fourth operation : 1 0 0 1 1 …
After the fifth operation  : 1 0 0 1 0 …

The later operations cannot change the condition of the fifth lamp any more. So the answer is 0.

翻译：之后的操作无法再改变第五个灯的状态，所以答案为0

 */
import java.util.Scanner;

public class Main2053 {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner sc = new Scanner(System.in);
        while(sc.hasNext()) {
            int n = sc.nextInt();
            int[] lamps = new int[n];
            int lampn = 0;
            for(int i = 1; i <= n; i++) {
                if(n % i == 0) {
                    if(lamps[n-1] == 0) {
                        lamps[n-1] = 1;
                    }else {
                        lamps[n-1] = 0;
                    }
                }lampn = lamps[n-1];
            }System.out.println(lampn);
        }
    }

}
